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vodka [1.7K]
3 years ago
11

How much work would be needed to raise the payload from the surface of the moon (i.e., x = r) to the "end of the universe"?

Physics
2 answers:
meriva3 years ago
6 0
Moon was rotation Earth or sun
it's own orbit . universe is not end
marysya [2.9K]3 years ago
3 0

Answer:

W(3R) - W(2R) = -PR² (1/(3R) - 1/(2R)) = PR/6

Explanation:

"Assume the weight move up at constant speed. With no net acceleration, the force applied is -(weight). Since the weight at height R is -P(R/x)² (minusbecause it's directed downward) the applied lifting force is P(R/x)², and the work done moving from x to x+dx is dW = P(R/x)² dx. Intetgrate this:"

W(x) = ?PR²/x² dx = -PR²/x + C

 The work done moving from x=R to x=2R is:

W(2R) - W(R) = -PR²(1/(2R) - 1/R) = PR/2

(b) The work done moving from 2R to 3R is:

W(3R) - W(2R) = -PR² (1/(3R) - 1/(2R)) = PR/6

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If two coils placed next to one another have a mutual inductance of 3.00 mH, what voltage (in V) is induced in one when the 2.50
hodyreva [135]

Answer:

-0.1875 V.

Explanation:

Using

E₂ = MdI₁/dt........................ Equation 1

Where E₂ = Voltage induced in the second coil, M = mutual inductance of both coil, dI₁ = change in current in the first coil, dt = change in time.

Given: M = 3.00 mH = 0.003 H, dI₁ = (0-2.50) = -2.5 A, dt = 40 ms = 0.04 s.

Substitute into equation 1

E₂ = 0.003(-2.5)/0.04

E₂ = -0.1875 V.

Hence the induced emf = -0.1875 V.

3 0
3 years ago
Anuja is holding a stuffed dog, with a mass of 0.30 kg, when Derek decides that he wants it and tries to pull it away from Abuja
igor_vitrenko [27]

Answer:

a = 3.33 m/s²

Explanation:

The horizontal acceleration of the dog can be found by using Newton's Second Law of Motion as follows:

F = ma

where,

F = Unbalanced force applied on the dog = 11 N - 10 N = 1 N

m = mass of the dog = 0.3 kg

a = horizontal acceleration of dog = ?

Therefore,

1 N = 0.3 kg(a)

a = 1 N/0.3 kg

<u>a = 3.33 m/s²</u>

8 0
3 years ago
g Drop the object again and carefully observe its motion after it hits the ground (it should bounce). (Consider only the first b
Anestetic [448]

Answer:

a) quantity to be measured is the height to which the body rises

b) weighing the body , rule or fixed tape measure

c)   Em₁ = m g h

d) deformation of the body or it is transformed into heat during the crash

Explanation:

In this exercise of falling and rebounding a body, we must know the speed of the body when it reaches the ground, which can be calculated using the conservation of energy, since the height where it was released is known.

a) What quantities must you know to calculate the energy after the bounce?

The quantity to be measured is the height to which the body rises, we assume negligible air resistance.

So let's use the conservation of energy

starting point. Soil

          Em₀ = K = ½ m v²

final point. Higher

          Em_f = U = mg h

         Em₀ = Em_f

         Em₀ = m g h₀

b) to have the measurements, we begin by weighing the body and calculating its mass, the height was measured with a rule or fixed tape measure and seeing how far the body rises.

c) We use conservation of energy

starting point. Soil

          Em₁ = K = ½ m v²

final point. Higher

          Em_f = U = mg h

         Em₁ = Em_f

         Em₁ = m g h

d) to determine if the energy is conserved, the arrival energy and the output energy must be compared.

There are two possibilities.

* that have been equal therefore energy is conserved

* that have been different (most likely) therefore the energy of the rebound is less than the initial energy, it cannot be stored in the possible deformation of the body or it is transformed into heat during the crash

7 0
2 years ago
An iron container has a mass of 200 g and contains 50 g of water @ 40°C. 50 g of ice @ -6°C are poured. Calculate the equilibr
malfutka [58]

Answer:

final equilibrium temperature of the system is ZERO degree Celcius

Explanation:

Hear heat given by water + iron = heat absorbed by ice

so here first we will calculate the heat given by water + iron

Q_1 = m_1s_2\Delta T_1 + m_2 s_2 \Delta T_1

Q_1 = (200)(0.450)(40 - T) + (50)(4.186)(40 - T)

now the heat absorbed by ice so that it will melt and come to the final temperature

Q_2 = m s \Delta T + mL + m s_{water}\Delta T'

Q_2 = 50(2.09)(0 + 6) + 50(335) + 50(4.186)(T - 0)

now we will have

17377 + 209.3T = 3600 - 90T + 8372 - 209.3T

17377 + 209.3T + 90T + 209.3T = 11972

T = -10.6

since it is coming out negative which is not possible so here the ice will not completely melt

so final equilibrium temperature of the system is ZERO degree Celcius

4 0
3 years ago
What are density conversions
shutvik [7]
\rho= \frac{kg}{m^3} = \frac{g}{cm^3} = \frac{g}{dm^3} \ and \ many \ others.
8 0
2 years ago
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